QPaperGen

We use the identity:
\[
\cos x + \cos y = 2 \cos\left(\frac{x + y}{2}\right) \cos\left(\frac{x - y}{2}\right)
\]

Now consider the left-hand side:
\begin{align*}
&2\cos\left(\frac{\pi}{13}\right)\cos\left(\frac{9\pi}{13}\right) + \cos\left(\frac{3\pi}{13}\right) + \cos\left(\frac{5\pi}{13}\right) \\
&= 2\cos\left(\frac{\pi}{13}\right)\cos\left(\frac{9\pi}{13}\right) + 2\cos\left(\frac{3\pi/13 + 5\pi/13}{2}\right)\cos\left(\frac{3\pi/13 - 5\pi/13}{2}\right) \\
&= 2\cos\left(\frac{\pi}{13}\right)\cos\left(\frac{9\pi}{13}\right) + 2\cos\left(\frac{4\pi}{13}\right)\cos\left(-\frac{\pi}{13}\right) \\
&= 2\cos\left(\frac{\pi}{13}\right)\cos\left(\frac{9\pi}{13}\right) + 2\cos\left(\frac{4\pi}{13}\right)\cos\left(\frac{\pi}{13}\right) \\
&= 2\cos\left(\frac{\pi}{13}\right)\left[\cos\left(\frac{9\pi}{13}\right) + \cos\left(\frac{4\pi}{13}\right)\right]
\end{align*}

Now apply the identity again:
\[
\cos\left(\frac{9\pi}{13}\right) + \cos\left(\frac{4\pi}{13}\right) = 2\cos\left(\frac{13\pi}{26}\right)\cos\left(\frac{5\pi}{26}\right)
\]

Since \(\cos\left(\frac{13\pi}{26}\right) = \cos\left(\frac{\pi}{2}\right) = 0\), we get:
\[
\cos\left(\frac{9\pi}{13}\right) + \cos\left(\frac{4\pi}{13}\right) = 0
\]

Therefore:
\[
2\cos\left(\frac{\pi}{13}\right) \cdot 0 = 0 \qquad \textbf{Hence, proved.}
\]